Is There a Power of 3 That Ends in 001?

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Discussion Overview

The discussion centers around the question of whether there exists a power of 3 that ends in 001. Participants explore mathematical reasoning related to modular arithmetic and the properties of powers of integers.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant proposes a proof involving the integers $3^1, 3^2, \ldots, 3^{1001}$ and the pigeonhole principle to show that there exist distinct integers $i$ and $j$ such that $3^i \equiv 3^j \pmod{1000}$.
  • The same participant concludes that this leads to $3^{j-1} \equiv 1 \pmod{1000}$, suggesting a resolution to the original question.
  • Another participant expresses interest in exploring what other odd numbers could replace 001, indicating a broader inquiry into the properties of powers of 3.
  • This second participant also mentions that the number of such odd numbers less than 1000 divides $\phi(1000)$, hinting at a connection to number theory.

Areas of Agreement / Disagreement

There is no consensus on the existence of other odd numbers that can replace 001, and the discussion remains open to exploration of this topic.

Contextual Notes

The discussion involves assumptions about modular arithmetic and the properties of the Euler totient function, which may not be fully elaborated or resolved.

mathmaniac1
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Prove that there exists a power of 3 that ends in 001.
 
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mathmaniac said:
Prove that there exists a power of 3 that ends in 001.
consider the integers:
$3^1,3^2,\ldots,3^{1001}$

Some two distinct integers of these must leave the same remainder mod $1000$.
So there exist distinct $i$ and $j$ such that $3^i\equiv 3^j\pmod{1000}$.
WLOG $i<j$. Thus $3^i(3^{j-i}-1)\equiv 0\pmod{1000}$
Thus $3^{j-1}\equiv 1\pmod{1000}$ and we are done.
 
Good work!So quick!
Now what odd numbers can we replace for 001?
 
Last edited:
mathmaniac said:
Good work!So quick!
Now what odd numbers can we replace for 001?
I don't know if there's an analytic way to enumerate all such numbers. But it can be shown then the number of such numbers which are less then 1000 divides $\phi(1000)$.
 

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